Sequence And Series Question 542
Question: If $ \log 2,\ \log (2^{n}-1) $ and $ \log (2^{n}+3) $ are in A.P., then n =
[MP PET 1998; Karnataka CET 2000; Pb. CET 2001]
Options:
A) 5/2
B) $ {\log_2}5 $
C) $ {\log_3}5 $
D) 3/2
Show Answer
Answer:
Correct Answer: B
Solution:
As, $ \log 2,\ \log (2^{n}-1) $ and $ \log (2^{n}+3) $ are in A.P. Therefore, $ 2\log (2^{n}-1)=\log 2+\log (2^{n}+3) $
$ \Rightarrow (2^{n}-5)(2^{n}+1)=0 $ As $ 2^{n} $ cannot be negative, hence $ 2^{n}-5=0 $
$ \Rightarrow 2^{n}=5 $ or $ n={\log_2}5 $ .
BETA
